📂Complex Anaylsis

Proof of the Residue Theorem

Theorem ¹

Let the analytic function $f: A \subset \mathbb{C} \to \mathbb{C}$ have a finite number of singularities $z_{1} , z_{2} , \cdots , z_{m}$ inside a simple closed path $\mathscr{C}$. Then, $$\int_{\mathscr{C}} f(z) dz = 2 \pi i \sum_{k=1}^{m} \text{Res}_{z_{k}} f(z)$$

Explanation

At first glance, the theorem might seem quite confusing. One has to calculate the integral, but instead of calculus-like computations, there’s talk about singularities and residues; it might well be perplexing. The theorem suggests that one could find the integral value simply by adding up residues. But could it really be that straightforward? Surprisingly, the answer is ‘yes’, thanks to the residue theorem.

This approach not only replaces integral calculations with other calculations but also makes many integrals that were impossible to solve feasible. Some integrals that couldn’t be handled with real numbers become relatively straightforward with the application of the residue theorem. Complex analysis hosts many important theorems, but the residue theorem, in particular, provides especially useful results and is thus essential to understand.

Proof

First, let’s break $\mathscr{C}$ into $m$ parts.

Generalized contraction sublemma for partitions: Assuming $f: A \subseteq \mathbb{C} \to \mathbb{C}$ is analytic at all points inside a simple closed path $\mathscr{C}$, excluding a finite number of points $z_{1} , z_{2}, \cdots z_{m}$ within a simply connected region that contains $\mathscr{C}$. Then, for a circle $\mathscr{C_k}$ centered at $z_{k}$ inside $\mathscr{C}$, $$\int_{\mathscr{C}} f(z) dz = \sum_{k=1}^{m} \int_{\mathscr{C}_{k}} f(z) dz$$

If we expand each $\mathscr{C}_{k}$ using a Laurent series, $$\int_{\mathscr{C}_{k}} f(z) dz = \int_{\mathscr{C}_{k}} \sum_{n = 0 }^{\infty} a_{nk} (z-z_{k}) ^{n} dz + \int_{\mathscr{C}_{k}} \sum_{n = 1 }^{\infty} { {b_{nk} } \over{ (z-z_{k}) ^{n} } } dz$$ by the Cauchy’s theorem, $$\int_{\mathscr{C}_{k}} f(z) dz = \int_{\mathscr{C}_{k}} \sum_{n = 1 }^{\infty} { {b_{nk} } \over{ (z-z_{k}) ^{n} } } dz$$ meanwhile, since $\int_{\mathscr{C}_{k}} {{1} \over {(z - z_{k})^n}} dz = \begin{cases} 2 \pi i & n = 1 \\ 0 & n \ge 2 \end{cases}$ by Cauchy’s integral formula, $$\int_{\mathscr{C}_{k}} f(z) dz = 2 \pi i b_{1k} = 2 \pi i \text{Res}_{z_{k}} f(z)$$ Therefore, $$\int_{\mathscr{C}} f(z) dz = 2 \pi i \sum_{k=1}^{m} \text{Res}_{z_{k}} f(z)$$

■

Notes

Especially in the proof, it is important to note that when $n=1$, all coefficients except for the residue $b_1k$, that is, $\displaystyle {{1} \over {z - z_{k}}}$, become $0$ and vanish. The residue theorem can be so beneficial that if one only studies its applications, it might be easy to forget why such results occur. Reminding oneself of the form of Cauchy’s integral formula and the Laurent series could signify a thorough understanding.